Problems

In a graph \(G\), we call a matching any choice of edges in \(G\) in such a way that all vertices have only one edge among chosen connected to them. A perfect matching is a matching which is arranged on all vertices of the graph.
Let \(G\) be a graph with \(2n\) vertices and all the vertices have degree at least \(n\) (the number of edges exiting the vertex). Prove that one can choose a perfect matching in \(G\).

A new customer comes to the hotel and wants a room. It happened today that all the rooms are occupied. What should you do?

Now imagine you got \(10\) new guests arriving to the completely full hotel. What should you do now?

The next day you have even harder situation: to the hotel, where all the rooms are occupied arrives a bus with infinitely many new customers. In the bus all the seats have numbers \(1,2,3...\) corresponding to all natural numbers. How to deal with this one?

Imagine you have \(2\) new guests arriving to the full hotel. How do you accommodate them?

What would you do about \(10000\) new guests arriving to the full hotel?

Imagine you have now a general finite number of new guests arriving to the full hotel. What do you do?

Today we will draw lots of pictures.

The subject is Topology. It is often called “rubber-sheet geometry" because while it is the study of shapes, topologists typically do not pay too much attention to rigid notions like angle and lengths. We have much more flexibility in topology. Some common words describing the operations here might include “gluing", “stretching", “twisting" and “inflating".

Although we will not define continuity, it is a more or less intuitive idea. Topological operations should be continuous. If you have a line segment, no amount of stretching, twisting or bending can make it into two disconnected segments.

In this sheet, we will look at basic counting problems. The fundamental principle is quite simple. If you have two independent choices to make, then the number of options for making both choices is calculated by multiplying the number of options for each choice.

An issue we frequently run into is that of overcounting. This means we count the same thing more than once. In the examples and problems today, you will see various ideas that we can use to correct for overcounting, or for avoiding it.

From the examples above, we see that we often need to pick \(k\) objects from \(n\) objects where the order of the \(k\) objects is ignored. The number of ways to pick them is notated with the special symbol \(\binom{n}{k}\), pronounced “\(n\) choose \(k\)". What’s a formula for \[\binom{n}{k}\]?